Inference for Numerical Data
Brian K. Liles
October 23, 2017
North Carolina Births
Exploratory Analysis
#load the nc data set into our workspace
download.file("https://www.openintro.org/stat/data/nc.RData", destfile="nc.RData")
load("nc.RData")Exercise 1
What are the cases in this data set? How many cases are there in our sample?
#dimensions of the nc data set
dim(nc)## [1] 1000 13The data set has 1000 observations and 13 variables
#view the head of the data set
head(nc)## fage mage mature weeks premie visits marital gained weight
## 1 NA 13 younger mom 39 full term 10 married 38 7.63
## 2 NA 14 younger mom 42 full term 15 married 20 7.88
## 3 19 15 younger mom 37 full term 11 married 38 6.63
## 4 21 15 younger mom 41 full term 6 married 34 8.00
## 5 NA 15 younger mom 39 full term 9 married 27 6.38
## 6 NA 15 younger mom 38 full term 19 married 22 5.38
## lowbirthweight gender habit whitemom
## 1 not low male nonsmoker not white
## 2 not low male nonsmoker not white
## 3 not low female nonsmoker white
## 4 not low male nonsmoker white
## 5 not low female nonsmoker not white
## 6 low male nonsmoker not white#summary
summary(nc)## fage mage mature weeks
## Min. :14.00 Min. :13 mature mom :133 Min. :20.00
## 1st Qu.:25.00 1st Qu.:22 younger mom:867 1st Qu.:37.00
## Median :30.00 Median :27 Median :39.00
## Mean :30.26 Mean :27 Mean :38.33
## 3rd Qu.:35.00 3rd Qu.:32 3rd Qu.:40.00
## Max. :55.00 Max. :50 Max. :45.00
## NA's :171 NA's :2
## premie visits marital gained
## full term:846 Min. : 0.0 married :386 Min. : 0.00
## premie :152 1st Qu.:10.0 not married:613 1st Qu.:20.00
## NA's : 2 Median :12.0 NA's : 1 Median :30.00
## Mean :12.1 Mean :30.33
## 3rd Qu.:15.0 3rd Qu.:38.00
## Max. :30.0 Max. :85.00
## NA's :9 NA's :27
## weight lowbirthweight gender habit
## Min. : 1.000 low :111 female:503 nonsmoker:873
## 1st Qu.: 6.380 not low:889 male :497 smoker :126
## Median : 7.310 NA's : 1
## Mean : 7.101
## 3rd Qu.: 8.060
## Max. :11.750
##
## whitemom
## not white:284
## white :714
## NA's : 2
##
##
##
## The following are categorical variables: mature, premie, marital, lowbirthweight, gender, habit, whitemom
Exercise 2
Make a side-by-side boxplot of habit and weight. What does the plot highlight about the relationship between these two variables?
boxplot(nc$habit,nc$weight, col="blue", main="Habit & Weight", xlab = "Birth Weight (lbs)", horizontal = TRUE)
#compare the means of the distributions using the following function to split the weight variable into the habit groups, then take the mean of each using the mean function
by(nc$weight, nc$habit, mean)## nc$habit: nonsmoker
## [1] 7.144273
## --------------------------------------------------------
## nc$habit: smoker
## [1] 6.82873Inference
Exercise 3
Check if the conditions necessary for inference are satisfied. Note that you will need to obtain sample sizes to check the conditions. You can compute the group size using the same by command above but replacing mean with length.
by(nc$weight, nc$habit, length)## nc$habit: nonsmoker
## [1] 873
## --------------------------------------------------------
## nc$habit: smoker
## [1] 126Exercise 4
Write the hypotheses for testing if the average weights of babies born to smoking and non-smoking mothers are different. H0: Undeniable evidence that babies born to smoking mothers will have an average birth weight H1: Undeniable evidence that babies born to smoking mothers will not have an average birth weight
#y is the response variable
#x is the explanatory variable as it splits the variable into two groups, smokers & non-smokers
#est mean
#ht hyptotheses test
inference(y=nc$weight, x=nc$habit, est = "mean", type = "ht", null=0, alternative="twosided", method="theoretical")## Warning: package 'BHH2' was built under R version 3.4.2## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862## Observed difference between means (nonsmoker-smoker) = 0.3155
##
## H0: mu_nonsmoker - mu_smoker = 0
## HA: mu_nonsmoker - mu_smoker != 0
## Standard error = 0.134
## Test statistic: Z = 2.359
## p-value = 0.0184
#Exercise 5 Change the type argument to “ci” to construct and record a confidence interval for the difference between the weights of babies born to smoking and non-smoking mothers.
#by default the function reports an interval for the (mean)nonsmoker - (mean)smoker, we change this using the order argument
inference(y=nc$weight, x=nc$habit, est="mean", type="ci", null=0, alternative = "twosided", method = "theoretical", order = c("smoker","nonsmoker"))## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_smoker = 126, mean_smoker = 6.8287, sd_smoker = 1.3862
## n_nonsmoker = 873, mean_nonsmoker = 7.1443, sd_nonsmoker = 1.5187
## Observed difference between means (smoker-nonsmoker) = -0.3155
##
## Standard error = 0.1338
## 95 % Confidence interval = ( -0.5777 , -0.0534 )On Your Own
Calculate a 95% confidence interval for the average length of pregnancies (weeks) and interpret it in context. Note that since you’re doing inference on a single population parameter, there is no explanatory variable, so you can omit the x variable from the function.
inference(y = nc$weeks, est = "mean", type = "ci", method = "theoretical")## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 95 % Confidence interval = ( 38.1528 , 38.5165 )Calculate a new confidence interval for the same parameter at the 90% confidence level. You can change the confidence level by adding a new argument to the function: conflevel = 0.90.
inference(y = nc$weeks, est = "mean", type = "ci", method = "theoretical", conflevel = 0.90)## Single mean
## Summary statistics:
## mean = 38.3347 ; sd = 2.9316 ; n = 998
## Standard error = 0.0928
## 90 % Confidence interval = ( 38.182 , 38.4873 )Conduct a hypothesis test evaluating whether the average weight gained by younger mothers is different than the average weight gained by mature mothers.
inference(y=nc$weight, x=nc$mature, est = "mean", type = "ht", null=0, alternative="twosided", method="theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_mature mom = 133, mean_mature mom = 7.1256, sd_mature mom = 1.6591
## n_younger mom = 867, mean_younger mom = 7.0972, sd_younger mom = 1.4855## Observed difference between means (mature mom-younger mom) = 0.0283
##
## H0: mu_mature mom - mu_younger mom = 0
## HA: mu_mature mom - mu_younger mom != 0
## Standard error = 0.152
## Test statistic: Z = 0.186
## p-value = 0.8526
Now, a non-inference task: Determine the age cutoff for younger and mature mothers. Use a method of your choice, and explain how your method works.
First, create two subsets based on the variable fage. Younger mothers could be all mothers less than 30 while the mature mothers would be 30 and above. Since this data is categorical one would use the inference code from exercise 4.
Pick a pair of numerical and categorical variables and come up with a research question evaluating the relationship between these variables. Formulate the question in a way that it can be answered using a hypothesis test and/or a confidence interval. Answer your question using the inference function, report the statistical results, and also provide an explanation in plain language.
Categorical data based on premie & whitemom H0: Prove that more premies are born to white mothers H1: Dispute the fact the more premies are born to white mothers
inference(y=nc$whitemom, x=nc$premie, est = "proportion", type = "ht", null=0, alternative="twosided", method="theoretical", success="white")## Response variable: categorical, Explanatory variable: categorical
## Two categorical variables
## Difference between two proportions -- success: white
## Summary statistics:
## x
## y full term premie Sum
## not white 228 56 284
## white 616 96 712
## Sum 844 152 996## Observed difference between proportions (full term-premie) = 0.0983
##
## H0: p_full term - p_premie = 0
## HA: p_full term - p_premie != 0
## Pooled proportion = 0.7149
## Check conditions:
## full term : number of expected successes = 603 ; number of expected failures = 241
## premie : number of expected successes = 109 ; number of expected failures = 43
## Standard error = 0.04
## Test statistic: Z = 2.47
## p-value = 0.0134
Numerical data based on premie & whitemom H0: Prove that heavier babies are boys H1: Dispute the fact the heavier babies are boys
inference(y=nc$weight, x=nc$gender, est = "mean", type = "ht", null=0, alternative="twosided", method="theoretical")## Response variable: numerical, Explanatory variable: categorical
## Difference between two means
## Summary statistics:
## n_female = 503, mean_female = 6.9029, sd_female = 1.4759
## n_male = 497, mean_male = 7.3015, sd_male = 1.5168## Observed difference between means (female-male) = -0.3986
##
## H0: mu_female - mu_male = 0
## HA: mu_female - mu_male != 0
## Standard error = 0.095
## Test statistic: Z = -4.211
## p-value = 0